Show that $C(I)^n$ is a Banach space under the following norm

Solution 1:

Let me start from scratch. Let $I=[0,1]$. We have $C(I)^n:=C(I,\mathbb{R}^n)$, the continuous functions from $I$ to $\mathbb{R}^n$. Each continuous function $I\to\mathbb{R}^n$ can be identified with a $n$-tuple of functions $I\to\mathbb{R}$, namely the coordinate functions; i.e. if $f\in C(I,\mathbb{R}^n)$, then there exist unique continous functions $f_1,\dots,f_n\in C(I,\mathbb{R})$ such that $f(t)=(f_1(t),\dots,f_n(t))$ for all $t\in I$. Conversely, given any $n$-tuple of continous functions $f_1,\dots,f_n\in C(I,\mathbb{R}))$, the function defined by $f(t):=(f_1(t),\dots,f_n(t))$ is a function of $C(I,\mathbb{R}^n)$. We thus have an identification of $C(I,\mathbb{R}^n)$ with $C(I)\times\dots\times C(I)$ ($n$-times) as sets and this identification is easily seen to be linear, so this is an isomorphism of vector spaces.

The norm defined on on $C(I,\mathbb{R}^n)$ is given by $\|f\|=\max_{i=1,\dots,n}\|f_i\|_\infty$, where $f=(f_1,\dots,f_n)$ and for $g\in C(I)$, $\|g\|_\infty:=\max_{x\in I}|g(x)|$. So let's break the problem in 2 easier problems:

Assuming that $(X_1,\|\cdot\|_1),\dots, (X_n,\|\cdot\|_n)$ are Banach spaces, consider their cartesian product $Y:=X_1\times \dots\times X_n$. This is trivially a vector space. Define the norm $\|(x_1,\dots,x_n)\|:=\max_{i=1,\dots,n}\|x_i\|_i$ for all $(x_1,\dots,x_n)\in Y$. Show that $(Y,\|\cdot\|)$ is also a Banach space (i.e. complete). I leave this as an exercise to you, I am pretty sure you can do it (do it for $n=2$ for simplicity, then induction proves it for a general $n$ as well)
Prove that $(C(I),\|\cdot\|_\infty)$ is a Banach space, i.e. complete. But this is standard and well-covered fact, see for example the accepted answer on this post Space of bounded continuous functions is complete (and use the fact that continous functions over a compact set are automatically bounded to apply it to your case).